Difference between revisions of "1985 AIME Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | Since <math>x_n=\frac{n}{x_{n-1}}</math>, <math>x_n \cdot x_{n - 1} = n</math>. Setting <math>n = 2, 4, 6</math> and <math>8</math> in this equation gives us respectively <math>x_1x_2 = 2</math>, <math>x_3x_4 = 4</math>, <math>x_5x_6 = 6</math> and <math>x_7x_8 = 8</math> so < | + | Since <math>x_n=\frac{n}{x_{n-1}}</math>, <math>x_n \cdot x_{n - 1} = n</math>. Setting <math>n = 2, 4, 6</math> and <math>8</math> in this equation gives us respectively <math>x_1x_2 = 2</math>, <math>x_3x_4 = 4</math>, <math>x_5x_6 = 6</math> and <math>x_7x_8 = 8</math> so <cmath>x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.</cmath> Notice that the value of <math>x_1</math> was completely unneeded! |
== See also == | == See also == | ||
Revision as of 14:12, 6 March 2014
Problem
Let 
, and for 
, let 
. Calculate the product 
.
Solution
Since , 
. Setting 
and 
in this equation gives us respectively 
, 
, 
and 
so ![\[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.\]](http://latex.artofproblemsolving.com/d/c/5/dc517b8bd93adf1e5d719d491779824034848b4b.png)
Notice that the value of 
was completely unneeded!
See also
| 1985 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
